Stochastic Processes in Dynamic Systems
ECE 272A
Midterm Exam-Solution
Instructor: Behrouz Touri
1. For each of the following statements, determine if the statement is true or false. If
you believe the statement is wrong, provide a counterexample, otherwise, prove the
statement.
a. If a linear system x˙ = A(t)x has a solution for some x(0) = x0 ∈ Rn, the solution is
always unique.
b. A dynamics x˙ = f(x) cannot have multiple marginally stable equilibrium.
Solution:
a. False.
Consider the one-dimensional linear system defined by x˙ = A(t)x, where
A(t) =
{
2
t ∀ t > 0
0 t = 0
and let x(0) = 0. For this linear system you can verify that x(t) = 0, for all t, is a
solution. Also, note that x(t) = t2 for all t ≥ 0 is another solution.
b. False.
The dynamics x˙ = f(x) where f(x) = 0 for all x ∈ R has multiple marginally stable
equilibrium (namely all the points on the real line).
2. A model for the unicycle’s movement on the plain whose linear velocity is v and angular
velocity is ω is given by
p˙ =
(
v cos(θ)
v sin(θ)
)
, θ˙ = ω, (1)
where p(t) ∈ R2 is the the Cartesian coordinates of unicycle and θ its orientation. Let
the input to this system be u =
(
v
ω
)
∈ R2.
(a) Let
x =
x1x2
x3
 =
 p1 cos(θ) + (p2 − 1) sin(θ)−p1 sin(θ) + (p2 − 1) cos(θ)
θ
 .
Write down the state-space formulation of this system using state variable x and
input variable u, i.e., find out f(x, u) such that x˙ = f(x, u).
1
Solution: By calculating x˙ and plugging in p˙ and θ˙, we have
x˙ =
 p˙1 cos(θ)− p1θ˙ sin(θ) + p˙2 sin(θ) + (p2 − 1)θ˙ cos(θ)−p˙1 sin(θ) + p1θ˙ cos(θ) + p˙2 cos(θ)− (p2 − 1)θ˙ sin(θ)
θ˙

=
v − p1ω sin(θ) + (p2 − 1)ω cos(θ)p1ω cos(θ)− (p2 − 1)ω sin(θ)
ω

=
v + x2ω−x1ω
ω
 .
(b) Show that the origin (x∗ = (0, 0, 0)T and u∗ = (0, 0)T ) is an equilibirum for the
dynamics in part (a). Find the linearization x˙ = Ax+Bu around this equilibrium.
Solution: Since
f(x∗, u∗) =
v∗ + x∗2ω∗−x∗1ω∗
ω∗
 =
00
0
 ,
the origin is an equilibrium. We have
∇xf(x, u) =
 0 ω 0−ω 0 0
0 0 0
 ,∇uf(x, u) =
1 x20 −x1
0 1
 .
Therefore, the linearization around the equilibrium is
x˙ = ∇xf(x∗, u∗)x+∇uf(x∗, u∗)u =
0 0 00 0 0
0 0 0
x+
1 00 0
0 1
u.
(c) Show that ω(t) = 1 = v(t) and p(t) =
(
sin(t)
1− cos(t)
)
and θ(t) = t is a solution
for the original dynamics (1). Find the linearization of the dynamics in Part (a)
around this trajectory.
Solution: We have
x =
 p1 cos(θ) + (p2 − 1) sin(θ)−p1 sin(θ) + (p2 − 1) cos(θ)
θ
 =
 0−1
t
 , u = (v
ω
)
=
(
1
1
)
.
Therefore,
x˙ =
d
dt
 0−1
t
 =
00
1
 .
2
This satisfies
x˙ =
v + x2ω−x1ω
ω
 =
1 + (−1)1−0 · 1
1
 =
00
1
 .
Hence this is a solution. The linearization around the trajectory is
x˙ = ∇xf(x∗(t), u∗(t))x+∇uf(x∗(t), u∗(t))u =
 0 1 0−1 0 0
0 0 0
x+
1 −10 0
0 1
u.
3. We say that a linear time-varying system x˙ = A(t)x(t) is preserving direction u 6= 0 ∈ Rn
if A(t)u = λ(t)u, where n×n matrix A(t) and λ(t) ∈ R are continuous functions of time
t ≥ 0.
(a) Find the solution of the system starting at the initial condition x(0) = αu for some
α ∈ R.
Solution: Suppose x(t) = g(t)u, where g(t) ∈ R, g(0) = α. Since x˙ = A(t)x(t), we
have
g˙(t)u = A(t)g(t)u = g(t)A(t)u = g(t)λ(t)u.
Since u 6= 0, g˙(t) = g(t)λ(t). Therefore,
g(t) = g(0) exp
( ∫ t
0
λ(t)dt
)
= α exp
( ∫ t
0
λ(t)dt
)
.
Now, since x(t) = αu exp
( ∫ t
0 λ(t)dt
)
satisfies x˙ = A(t)x(t) and the initial condition
and A(t) is a continuous function, it is the solution to the ODE.
(b) Show that for all α 6= 0 in part (a), we have limt→∞ ‖x(t)‖ = 0, if and only if∫∞
0 λ(t)dt = −∞.
Solution: We have
‖x(t)‖ = ‖αu exp ( ∫ t
0
λ(t)dt
)‖ = ‖αu‖ exp ( ∫ t
0
λ(t)dt
)
,
and ‖αu‖ > 0 since α 6= 0 and u 6= 0. Therefore, if ∫∞0 λ(t)dt = −∞, then
lim
t→∞ ‖x(t)‖ = ‖αu‖ limt→∞ exp
( ∫ t
0
λ(t)dt
)
= 0.
Also, if limt→∞ ‖x(t)‖ = 0, then
lim
t→∞ exp
( ∫ t
0
λ(t)dt
)
=
1
‖αu‖ limt→∞ ‖x(t)‖ = 0,
which implies
∫∞
0 λ(t)dt = −∞.
3
4. Consider the dynamics:
x˙ =
(−1 1
−2 −3
)
x+
(
0
1
)
u(t).
a. Find the solution for the unforced system with the initial condition x(0) = (−1, 1)T .
Solution: By eigenvalue decomposition, we have(−1 1
−2 −3
)
=
(−1 + i −1− i
2 2
)(−2− i 0
0 −2 + i
)(−i
2
1−i
4
i
2
1+i
4
)
.
Therefore, the solution is
x(t) = eAtx(0)
=
(−1 + i −1− i
2 2
)(
e(−2−i)t 0
0 e(−2+i)t
)(−i
2
1−i
4
i
2
1+i
4
)(−1
1
)
=
( −12e(−2−i)t − 12e(−2+i)t
1+i
2 e
(−2−i)t + 1−i2 e
(−2+i)t
)
=
( −e−2t cos t
e−2t cos t+ e−2t sin t
)
.
Solution: By Laplace transform,
X(t) = (sI −A)−1x(0)
=
1
s2 + 4s+ 5
(
s+ 3 1
−2 s+ 1
)
x(0)
=
1
(s+ 2)2 + 1
(−s− 2
s+ 3
)
.
By Inverse Laplace transform,
x(t) =
( −e−2t cos t
e−2t cos t+ e−2t sin t
)
.
b. Determine whether the origin for the system is marginally stable, asymptotically
stable, or unstable. Find the stability properties of the (equilibrium) origin for the
unforced system.
Solution: Let λ1 = −2 − i, λ2 = −2 + i. Since Re[λi] < 0 for i = 1, 2, the origin is
asymptotically (and exponentially) stable.
c. Show that if the input u(t) is an integrable and uniformly bounded function, i.e.,
|u(t)| ≤ β for some β > 0 and all t ≥ 0, then the trajectory is also uniformly
bounded, i.e., ‖x(t)‖ ≤ βˆ for some βˆ > 0 and all t ≥ 0.
Solution: The solution is
x(t) = eAtx(0) +
∫ t
0
eA(t−τ)
(
0
1
)
u(τ)dτ.
4
Therefore,
‖x(t)‖ ≤ ‖eAtx(0)‖+
∥∥∥∥∫ t
0
eA(t−τ)
(
0
1
)
u(τ)dτ
∥∥∥∥
≤ ‖eAtx(0)‖+
∫ t
0
∥∥∥∥eA(t−τ)(01
)
u(τ)
∥∥∥∥dτ
≤ ‖eAtx(0)‖+
∫ t
0
‖eA(t−τ)‖
∥∥∥∥(01
)∥∥∥∥|u(τ)|dτ
≤ ‖eAtx(0)‖+
∫ t
0
‖eA(t−τ)‖ · 1 · βdτ
= ‖eAtx(0)‖+ β
∫ t
0
‖eλ1(t−τ)P1 + eλ2(t−τ)P2‖dτ
≤ ‖eAtx(0)‖+ β‖P1‖
∫ t
0
e−2(t−τ)dτ + β‖P2‖
∫ t
0
e−2(t−τ)dτ
= ‖eAtx(0)‖+ β(‖P1‖+ ‖P2‖)1− e
−2t
2
,
where A = λ1P1 + λ2P2 for some P1, P2. Since the first term converges to 0 and the
second is bounded for all t ≥ 0, ‖x(t)‖ is uniformly bounded.
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